RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
Maxima and Minima Excercise: 17.3
Maxima and Minima excercise 17.3 question 1 (iii)
Maxima and Minima Excercise 17.3 question 1 (v)
Maxima and Minima exercise 17.3 question 1 (vi)
Maxima and Minima excercise 17.3 question 1 (ix)Answer:
Maxima and Minima exercise 17.3 question 2 (i)
Maxima and Minima exercise 17.3 question 3
Maxima and Minima excercise 17.3 question 4
Maxima and Minima exercise 17.3 question 8
Benefits of picking RD Sharma Arithmetic Solutions from Career360 include:
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.3

Maxima and Minima excercise 17.3 question 1 (i)

Answer:

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) < 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value
Given:
$f(x)=x^{4}-62 x^{2}+120 x+9$
Explanation:
It is given that
$\begin{aligned} &f(x)=x^{4}-62 x^{2}+120 x+9 \\ &f^{\prime}(x)=4 x^{3}-62 * 2 x+120 \end{aligned}$
To find critical values ,
$\begin{aligned} &f^{\prime}(x)=0 \\ &4 x^{3}-124 x+120=0 \\ &4\left(x^{3}-31 x+30\right)=0 \end{aligned}$
On solving,
critical values of x are 5,1,-6.
Now,
$f^{\prime \prime}(x)=12 x^{2}-1$
At x=5,
$\begin{aligned} f^{\prime \prime}(5) &=12(5)^{2}-124 \\ &=176>0 \end{aligned}$
So, x=5 is point of local minima.
At x=1,
$\begin{aligned} f^{\prime \prime}(1) &=12(1)^{2}-124 \\ &=-112<0 \end{aligned}$
So, x=1 is point of local maxima.
At x=-6,
$\begin{aligned} f^{\prime \prime}(-6) &=12(-6)^{2}-124 \\ &=308>0 \end{aligned}$

So, x=-6 is a point of local minima.
The local max. value at x=1 is
$\begin{aligned} f(1) &=(1)^{4}-62(1)^{2}+120(1)+9 \\ &=68 \end{aligned}$
And local min. value at x=5 & x=-6 are,
$\begin{aligned} f(5)=&(5)^{4}-62(5)^{2}+120(5)+9 \\ &=-316 \\ f(-6) &=(-6)^{4}-62(-6)^{2}+120(-6)+9 \\ &=-1647 \end{aligned}$
Hence, point of local maxima is 1 & its maximum value is 68.
Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.

Maxima and Minima excercise 17.3 question 1 (ii)
Answer:

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-6 x^{2}+9 x+15$
Explanation:
We have,

$\begin{aligned} f(x) &=x^{3}-6 x^{2}+9 x+15 \\ f^{\prime}(x) &=3 x^{2}-6.2 x+9 \\ &=3 x^{2}-12 x+9 \\ \therefore f^{\prime}(x) &=3\left(x^{2}-4 x+3\right) \\ \& f^{\prime \prime}(x) &=3(2 x-4) \\ &=6 x-12 \end{aligned}$
To find maxima and minima.
$\begin{aligned} &\qquad f^{\prime}(x)=0 \\ &3\left(x^{2}-4 x+3\right)=0 \\ &\quad x^{2}-4 x+3=0 \\ &X=3 \text { or } x=1 \\ &\text { At } x=3, \\ &f^{\prime \prime}(3)=6(3)-12 \\ &=6>0 \end{aligned}$
At x = 3,
f’’(3) = 6(3)-12
=6>0
So, x = 3 is point of local minima
At x = 1,
f’’(1) = 6(1)-12
=-6<0
So, x = 1 is point of local maxima
Now, local maximum value at x= 1 is
$\begin{aligned} &f(1)=(1)^{3}-6(1)^{2}+9(1)+15 \\ &f(1)=19 \end{aligned}$

And local min. value at x=3 is
$\begin{aligned} &f(3)=(3)^{3}-6(3)^{2}+9(3)+15 \\ &f(3)=15 \end{aligned}$
Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

Maxima and Minima excercise 17.3 question 1 (iii)

Answer:
Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is -4
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x-1)(x+2)^{2}$
Explanation:
We have,
$\begin{aligned} f(x) &=(x-1)(x+2)^{2} \\ f^{\prime}(x) &=(x+2)^{2}(1)+(x-1) 2(x+2) \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=(x+2)(x+2+2 x-2) \\ &=(x+2)(3 x) \\ f^{\prime \prime}(x) &=3 x(1)+(x+2) 3 \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=6 x+6 \end{aligned}$
For max and min, f’(x)=0,
$\begin{aligned} &(x+2)(3 x)=0 \\ &x=0 \& x=-2 \end{aligned}$
At x=0,
$f^{\prime \prime}(0)=6(0)+6=6>0$
So, x= 0 is point of local minima
$f^{\prime \prime}(0)=6(0)+6=6>0$
At x=-2,
$\begin{aligned} f^{\prime \prime}(0) &=6(-2)+6 \\ &=-6<0 \end{aligned}$
So, x = -2 is point of local maxima
So, local max. value at x=-2 is
f(-2)=(-2-1)(-2+2)^{2}=0
And local min. value at x= 0 is
$f(0)=(0-1)(0+2)^{2}=-4$
Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.

Maxima and Minima exercise 17.3 question 1 (iv)

Answer:

Point of local maxima is 2 and it’s local maximum value is $\frac{1}{2}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f"(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0$
Explanation:
We have,
$\begin{gathered} f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0 \\ \therefore f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{4}{x^{3}} \\ f^{\prime \prime}(x)=\frac{4}{x^{2}}-\frac{12}{x^{4}} \end{gathered}$

For maxima and minima, f’(x)=0
$\begin{array}{r} -\frac{2}{x^{2}}+\frac{4}{x^{3}}=0 \\ -\frac{2(x-2)}{x^{3}}=0 \\ x=2 \end{array}$
At x=2,
$\begin{aligned} f^{\prime \prime}(2) &=\frac{4}{(2)^{3}}-\frac{12}{(2)^{2}} \\ &=\frac{4}{8}-\frac{12}{16} \\ &=\frac{1}{2}-\frac{3}{4} \\ &=-\frac{1}{4}<0 \end{aligned}$
So, x=2 is point of local maxima.
So, local max. value at x= 2 is
$\begin{aligned} f(2) &=\frac{2}{2}-\frac{2}{(2)^{2}} \\ &=1-\frac{2}{4}=\frac{1}{2} \end{aligned}$
Hence, point of local maxima is 2 & its max. value is $\frac{1}{2}$

Maxima and Minima Excercise 17.3 question 1 (v)

Answer:
Point of local minima value is x=-1 and it’s local minimum value is $-\frac{1}{e^{.}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) > 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given: $f(x)=x e^{\mathrm{x}}$
Explanation:
We have,
$\begin{aligned} &f(x)=x e^{\mathrm{x}} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \ldots \text { using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+1)+e^{\mathrm{x}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+2) \end{aligned}$
Now, for maxima & minima, f’(x)=0
$\begin{aligned} e^{\mathrm{x}}(x+1) &=0 \\ x+1 &=0 \\ x &=-1 \end{aligned}$
at x=-1,
$f^{\prime \prime}(-1)=e^{-1}(-1+2)=\frac{1}{e}$
X = -1 is point of local minima.
So, local min. value at x= 1 is
$f(-1)=-1 e^{-1}=-1 / e$
Thus, point of local minima is -1 & its min value is $-\frac{1}{e}$

Maxima and Minima exercise 17.3 question 1 (vi)

Answer:
Point of local minima value is 2 and it’s local minimum value is 2.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $\mathrm{c}_{1}$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$\frac{x}{2}+\frac{2}{x}, \underline{x}>0$
Explanation:
We have,
$\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned}$
For local maxima or minima, we have $f^{\prime}(x)=0$
$\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}$
since, $x>0$
$x = 2$
Now, at x=2 ,
$f(2)=\frac{2}{2}+\frac{2}{2}=2$
Thus, its min. value at 2 is 2.

Maxima and Minima Excercise 17.3 question 1 (vii)
Answer:

Point of local minima value is $-\frac{7}{4}$ and it’s local minimum value is $-\frac{3}{(4)^{\frac{4}{3}}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x+1)(x+2)^{\frac{1}{3}}, x>=-2$
Explanation:
we have ,
$\begin{aligned} &f(x)=(x+1)(x+2)^{-} \\ &f^{\prime}(x)=(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{\left(-\frac{2}{3}\right)} \ldots \text{ using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=\frac{2}{3(x+2)^{2 / 3}}-\frac{2}{9}(x+1)(x+2)^{\left(-\frac{5}{3}\right)} \ldots \text{ using }\frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \end{aligned}$
For maxima & minima, f’(x)=0
$\begin{aligned} &(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\frac{2}{3}}=0 \\ &\frac{1}{3}(x+1)=-(x+2)^{\frac{1}{3}}(x+2)^{\frac{2}{3}} \\ &\frac{1}{3}(x+1)=-(x+2) \\ &x+1=-3 x-6 \\ &x=-\frac{7}{4} \end{aligned}$
At x= $-\frac{7}{4}$
$\begin{aligned} f^{\prime \prime}\left(-\frac{7}{4}\right) &=\frac{2}{3}\left(-\frac{7}{4}+2\right)^{\frac{-2}{3}}-\frac{2}{9}\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\left(-\frac{5}{3}\right)} \\ &=\frac{2}{3}\left(\frac{1}{4}\right)^{-\frac{2}{3}}+\frac{1}{18}\left(\frac{1}{4}\right)^{-\frac{5}{3}}>0 \end{aligned}$
So , $x=-\frac{7}{4}$ is a point of local minima
Now at $x=-\frac{7}{4}$
$\begin{aligned} f\left(-\frac{7}{4}\right) &=\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\frac{1}{3}} \\ &=\left(-\frac{3}{4}\right)\left(\frac{1}{4}\right)^{\frac{1}{3}} \\ &=-\frac{3}{4^{\frac{4}{3}}} \end{aligned}$
Thus, point of local minima is $-\frac{7}{4}$ & its value is $-\frac{3}{4^{\frac{4}{3}}}$

Maxima and Minima exercise 17.3 question 1 (viii)

Answer:
Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is -16
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5$
Explanation:
We have,
$\begin{aligned} &f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5 \\ &f^{\prime}(x)=\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=-\frac{x}{\sqrt{32-x^{2}}}-\left(\frac{2 x \sqrt{32-x^{2}}+\frac{x^{3}}{\sqrt{32-x^{2}}}}{\left(32-x^{2}\right)}\right) \ldots u \sin g \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \end{aligned}$
Now for local minima and local maxima
$\begin{aligned} &f^{\prime}(x)=0 \\ &\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}=0 \\ &32-x^{2}=x^{2} \\ &x^{2}=16 \\ &x=\pm 4 \end{aligned}$
At x=4,
$\begin{aligned} f^{\prime \prime}(4) &=\frac{-4}{\sqrt{32-4^{2}}}-\left[\frac{\left(8\left(32-4^{2}\right)+4^{3}\right)}{\left(32-4^{2}\right)\left(\sqrt{32-4^{2}}\right.}\right] \\ &=-1-\frac{192}{64} \\ &=-3<0 \end{aligned}$
So, x=4 is a point of local maxima.
At x=4,
$\begin{aligned} f(4) &=4 \sqrt{32-(4)^{2}} \\ &=4 \sqrt{32-16} \\ &=16 \end{aligned}$
At x=-4,
$\begin{aligned} f^{\prime \prime}(-4) &=-\frac{4(-4)}{\sqrt{32-(-4)^{2}}}-\left[\frac{8\left(32-4^{2}\right)-(-4)^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right] \\ &=0+2=3>0 \end{aligned}$
So, x=-4 is the point of local minimum.
At x=-4
$\begin{aligned} f(x) &=(-4) \sqrt{32-(-4)^{2}} \\ &=-4 \sqrt{32-16}=-16 \end{aligned}$
Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

Maxima and Minima excercise 17.3 question 1 (ix)
Answer:

Point of local maxima value is $\frac{a}{3}$ and it’s local maximum value is $\frac{4 a^{3}}{27}.$ Also point of local minima is a and it’s local minimum value is 0.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(\mathrm{c}_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(\mathrm{c}_{2}\right)<0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0$
Explanation:
We have,
$\begin{aligned} &f(x)=x^{3}-2 a x^{2}+a^{2} x \\ &f^{\prime}(x)=3 x^{2}-4 a x+a^{2} \\ &f^{\prime \prime}(x)=6 x-4 a \end{aligned}$
For maxima and minima f’(x) =0
$\begin{aligned} &3 x^{2}-4 a x+a^{2}=0 \\ &x=\frac{-(-4 a) \pm \sqrt{(-4 a)^{2}-4 \times 3 \times a^{2}}}{2 \times 3} \\ &\quad=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{6} \\ &=\frac{4 a \pm 2 a}{6} \\ &x=a \text { or } x=\frac{a}{3} \end{aligned}$
At x= a,
$\begin{aligned} f^{\prime \prime}(a) &=6(a)-4 a \\ &=2 a>0 \end{aligned}$
So, x=a is point of local minima
Now at x=a,
$\begin{aligned} f(a) &=a^{3}-2 a(a)^{2}+a^{2}(a) \\ &=a^{3}-2 a^{3}+a^{3}=0 \end{aligned}$
Also, at $\mathrm{x}=\frac{a}{3}$
$\begin{aligned} f\left(\frac{a}{3}\right) &=\left(\frac{a}{3}\right)^{2}-2 a\left(\frac{a}{3}\right)+a^{2}\left(\frac{a}{3}\right) \\ &=\frac{4 a^{3}}{27} \end{aligned}$
Thus, point of local maxima is $\frac{a}{3}$ & its max. value is $\frac{4 a^{3}}{27}$ & point of local minima is a & its value is 0.

Maxima and Minima excercise 17.3 question 1 (x)

Answer:
Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is -2a
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x+\frac{a^{2}}{x}, a>0$
Explanation:
We Have,
$\begin{aligned} &f(x)=x+\frac{a^{2}}{x} \\ &f^{\prime}(x)=1-\frac{a^{2}}{x^{2}} \\ &f^{\prime \prime}(x)=\frac{2 a^{2}}{x^{3}} \end{aligned}$
For maxima and minima ,f’(x)=0
$\begin{aligned} 1-\frac{a^{2}}{x^{2}} &=0 \\ x &=\pm a \end{aligned}$

Maxima and minima exercise 17.3 question 1 (xi)

Answer:

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.
Hint:
Putting x=1,-1in f''(x)
Given:
$\begin{aligned} &f(x)=x \sqrt{2-x^{2}} \\ &-\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}$
Explanation:
Differentiating f(x) with respect to x
$f^{\prime}(x)=\frac{d}{d x}\left(x \sqrt{2-x^{2}}\right)$
$\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{g}(\mathrm{x})\}+\mathrm{g}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}+\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{-}(1) \end{aligned}$
$\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{df}(\mathrm{x})^{\mathrm{n}}}{\mathrm{dx}}=\mathrm{nf}(\mathrm{x})^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x}) \text { and }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})]=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x})) \pm \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{g}(\mathrm{x}))\\ &\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2-\mathrm{x}^{2}\right)^{1 / 2}\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2-1} \frac{d}{d x}\left(2-x^{2}\right)\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2}(-2 x)\\ &=-\frac{x}{\sqrt{2-x^{2}}} \end{aligned}$
$\begin{aligned} &\therefore \operatorname{From}(1), \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}\left(-\frac{\mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}\right)+\sqrt{2-\mathrm{x}^{2} \times 1} \\ &=\frac{-\mathrm{x}^{2}+2-\mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \\ &=\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \end{aligned}$
$\begin{aligned} &\text { Put, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=0\\ &\frac{2-2 x^{2}}{\sqrt{2-x^{2}}}=0\\ &\Rightarrow 2-2 x^{2}=0\\ &\Rightarrow 2 \mathrm{x}^{2}=2\\ &\Rightarrow \mathrm{x}^{2}=1\\ &\Rightarrow x=\pm 1\\ &\text { if } x \neq \sqrt{2} \text { and } x \neq-\sqrt{2} \end{aligned}$
These x=1 and x=-1 we the possible point of local maxima and minima
Differentiating f’(x) with respect to x
$\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}}\right) \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(2-2 \mathrm{x}^{2}\right)-\left(2-2 \mathrm{x}^{2}\right) \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}}{\left(2-\mathrm{x}^{2}\right)} \end{aligned}$
Using the formula,
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}-\mathrm{u} \frac{\mathrm{d}}{\mathrm{ax}} \mathrm{v}}{\mathrm{v}^{2}} \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \times(-4 \mathrm{x})-\frac{\left(2-2 \mathrm{x}^{2}\right) 1}{2}\left(2-\mathrm{x}^{2}\right)^{-1 / 2}(-2 \mathrm{x})}{\sqrt{2-\mathrm{x}^{2}} 2} \quad \text { (Applying chain rule) } \\ &=\frac{-4 \mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}+\frac{\mathrm{x}\left(2-2 \mathrm{x}^{2}\right)}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-4 \mathrm{x}\left(2-\mathrm{x}^{2}\right)+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-8 \mathrm{x}+4 \mathrm{x}^{3}+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \end{aligned}$
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=1 \\ &\mathrm{f}^{\prime \prime}(1)=\frac{-6(1)+2(1)^{3}}{\left(2-(1)^{2}\right)^{3 / 2}} \\ &=\frac{-6+2}{(1)^{3 / 2}}=-4<0 \end{aligned}$

So x=1 is local maxima and its maximum value f(1)=1

$\begin{aligned} &f^{\prime \prime}(x) \text { at } x=-1 \\ &f^{\prime \prime}(-1)=\frac{-6(-1)+(2)(-1)^{3}}{\left(2-(-1)^{2}\right)^{3 / 2}} \\ &=\frac{6-2}{(\sqrt{1})^{3}}=4>0 \end{aligned}$

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

Maxima and Minima Excercise 17.3 question 1 (xii)

Answer:
x= $\frac{3}{4}$ is a point of local maxima and its maximum value is 5/4.
Hint:
Using chain rule of Differentiation
Given:
$\begin{aligned} &\mathrm{f}(\mathrm{x})=\mathrm{x}+\sqrt{1-\mathrm{x}} \\ &\mathrm{x} \leq 1 \end{aligned}$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\sqrt{1-\mathrm{x}}) \\ &=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{1 / 2} \\ &=1+\frac{1}{2}(1-\mathrm{x})^{\frac{1}{2}-1}(-1) \end{aligned}$ [ Using Chain Rule Diffrentiation ]
$\begin{aligned} &=1-\frac{1}{2 \sqrt{1-x}} \\ &=\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}} \\ &\text { Put } f^{\prime}(x)=0 \\ &\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}}=0 \\ &\Rightarrow 2 \sqrt{1-x}-\mid 1=0 \text { if } x \neq 1 \\ &\Rightarrow \sqrt{1-x}=\frac{1}{2} \end{aligned}$ [Squaring Both Sides]
$\begin{aligned} &\text { i. } e 1-x=\frac{1}{4} \\ &x=\frac{3}{4} \end{aligned}$
Thus , $x=\frac{3}{4}$ is the point of local maxima and minima
Differentiating f’ with respect to x
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(1-\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=0-\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{-1 / 2}\\ &=\frac{1}{2} \times-\frac{1}{2}(1-x)^{-3 / 2}(-1)\\ &\text { (Using chain rule) }\\ &=-\frac{1}{4(1-x)^{\frac{3}{2}}} \end{aligned}$ $\begin{aligned} &\text { Put } \mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=\frac{3}{4} \\ &\mathrm{f}^{\prime \prime}\left(\frac{3}{4}\right)=\frac{-1}{4\left(1-\frac{3}{4}\right)^{3 / 2}} \\ &=\frac{-1}{4 \times \frac{1}{2 \sqrt{2}}} \\ &=-\frac{\sqrt{2}}{2}<0 \end{aligned}$

So $x = \frac{3}{4}$ is a point of local maxima
The local maximum value at f(x)
$\begin{aligned} &\mathrm{f}\left(\frac{1}{2}\right)=\frac{3}{4}+\sqrt{1-\frac{3}{4}} \\ &=\frac{3}{4}+\frac{1}{2}=\frac{3+2}{4}=\frac{5}{4} \end{aligned}$

Maxima and Minima exercise 17.3 question 2 (i)

Answer:

$x=2$ is the local minima and the local minimum value of $f(x)$ at $x=2$ is 0.
$x=\frac{4}{3}$ is the local maxima and the local maximum value of $f(x)$ at $x=\frac{4}{3}$ is $x=\frac{4}{27}$ .

Hint:
Using chain rule
Given:
$f(x)=(x-1)(x-2)^{2}$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\mathrm{x}-1)(\mathrm{x}-2)^{2}\right) \\ &=(\mathrm{x}-1) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-2)^{2}+(\mathrm{x}-2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1) \\ &=2(\mathrm{x}-1)(\mathrm{x}-2)+(\mathrm{x}-2)^{2} \quad \text { [Using chain rule] } \\ &=2\left(\mathrm{x}^{2}-2 \mathrm{x}-\mathrm{x}+2\right)+\left(\mathrm{x}^{2}-4 \mathrm{x}+4\right) \\ &=2 \mathrm{x}^{2}-4 \mathrm{x}-2 \mathrm{x}+4+\mathrm{x}^{2}-4 \mathrm{x}+4 \\ &=3 \mathrm{x}^{2}-10 \mathrm{x}+8 \end{aligned}$
$\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &3 x^{2}-10 x+8=0 \\ &x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{(10)^{2}-4 \cdot 3 \cdot 8}}{2 \cdot(3)} \\ &=\frac{+10 \pm \sqrt{100-96}}{6} \\ &=\frac{-10 \pm 2}{6} \end{aligned}$
$\begin{aligned} &\mathrm{x}=\frac{+10+2}{6}\\ &=2\\ &\text { And }\\ &x=\frac{+10-2}{6}\\ &=\frac{4}{3} \end{aligned}$

Differentiating $f(x)$ with respect to $x$
$\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-10 \mathrm{x}+\mathrm{i}\right) \\ &=6 \mathrm{x}-10 \end{aligned}$
When put x = 2 in f’’(x)
$\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}$
So x=2 is the local Minima
The Local Minimum value of f(x) at x =2 is 0
Put $x=\frac{4}{3} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})$
$\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}$
The local maximum value of $f(x)$ at $x=\frac{4}{3}$ is $x=\frac{4}{27}$ .

Maxima and Minima excercise 17.3 question 2 (iii)

Answer:
x=1 is point of inflexion
x=-1 is the point of local minimum & $x=\frac{1}{5}$ is the point of local maximum
Hint:
Using chain rule of derivative
Given:
$f(x)=-(x-1)^{3}(x+1)^{2}$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+1)^{2}+(\mathrm{x}+1)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^{3}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \cdot 2(\mathrm{x}+1)+3(\mathrm{x}+1)^{2}(\mathrm{x}-1)^{2}\right] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)[2 \mathrm{x}-2+3 \mathrm{x}+3] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)(5 \mathrm{x}+1) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+\mathrm{x}+5 \mathrm{x}+1\right) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}+6 \mathrm{x}^{3}+\mathrm{x}^{2}-10 \mathrm{x}^{3}-12 \mathrm{x}^{2}-2 \mathrm{x}+5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}-4 \mathrm{x}^{3}-6 \mathrm{x}^{2}+4 \mathrm{x}+1\right) \end{aligned}$
$\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &-1(x-1)^{2}(x+1)(5 x+1)=0 \\ &(x-1)^{2}(x+1)(5 x+1)=6 \\ &\Rightarrow(x-1)^{2}=0 \\ &x-1=0 \\ &x=1 \\ &x+1=0, \quad 5 x+1=0 \\ &x=-1, \quad x=-\frac{1}{5} \end{aligned}$
Thus $x=1$ and $x=-1$ and $x=-\frac{1}{5}$ are the possible points of local minima and maxima
Differentiating f’(x) with respect to x $\begin{aligned} &f^{\prime \prime}(x)=\frac{d}{d x}\left[-\left(5 x^{4}-4 x^{3}-6 x^{2}+4 x+1\right)\right] \\ &=-\left(20 x^{3}-12 x^{2}-12 x+4\right) \\ &=-20 x^{3}+12 x^{2}+12 x-4 \\ &\text { when } x=1 \\ &\begin{aligned} f^{\prime \prime}(1) &=-20(1)^{3}+12(1)^{2}+12(1)-4 \\ &=-20+12+12-4=0 \end{aligned} \end{aligned}$

Thus this test is fail as x=1 is point of inflexion
when $x=-1$
$\begin{aligned} f^{\prime \prime}(-1) &=-20(-1)^{3}+12(-1)^{2}+12(-1)-4 \\ &=20+12-12-4=16>0 \end{aligned}$

So , x = -1 is a point of local minimum

When $x=-\frac{1}{5}$

$\begin{aligned} f\left(-\frac{1}{5}\right)=&-20\left(-\frac{1}{5}\right)^{3}+12\left(-\frac{1}{5}\right)^{2}+12\left(-\frac{1}{5}\right)-4 \\ &=-336 / 125<0 \end{aligned}$
So, x=-1/5 is the point of local maximum

Maxima and Minima exercise 17.3 question 3

Answer:
$a = -\frac{2}{3}$
$\mathrm{b}=-\frac{1}{6}$
Hint:
Put? $\frac{d y}{d x}=0 \text { at } x=1,2$
Given:
$y=\operatorname{alog} x+b x^{2}+x$
Explanation:
Differentiating y with respect to x
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{~b} \mathrm{x}+1 \\ &\quad=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1 \\ &\text { At } \mathrm{x}=1 \mid \\ &\begin{aligned} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) &=\frac{\mathrm{a}}{1}+2 \mathrm{~b}(1)+1 \\ &=\mathrm{a}+2 \mathrm{~b}+1-(1) \end{aligned} \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &\frac{d y}{d x}=\frac{a}{2}+2(b) \cdot 2+1 \\ &=\frac{a}{2}+4 b+1-(2) \end{aligned}$
Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2
Then equation (1) and (2) become
$\begin{aligned} &a+2 b+1=0-(3)\\ &\frac{a}{2}+4 b+1=0\\ &a+8 b+2=0 \quad-(4)\\ &\text { [Multiplying both sides by 2] } \end{aligned}$
Solving , (3) and (4) , we get
$\begin{array}{r} 6 b-1=0 \\ b=-\frac{1}{6} \end{array}$
Putting $B = -1/6 in (3) , we get$
$\begin{aligned} &a-\frac{1}{3}+120 \\ &a=-\frac{2}{3} \end{aligned}$

Maxima and Minima excercise 17.3 question 4

Answer:
We need to prove that x = e is local maxima
Given:
The given function $\frac{\log x}{x}$
Explanation:
$\text { Let } y=\frac{\log x}{x}$
Then
$\begin{aligned} \frac{d y}{d x} &=\frac{x \cdot \frac{d}{d x}(\log x)-\log (x) \frac{d}{d x}(x)}{x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{x \times \frac{1}{x}-\log x}{x^{2}} \\ &=\frac{1-\log x}{x^{2}} \end{aligned}$
Put,
$\begin{aligned} &\frac{d y}{d x}=0 \\ &\frac{1-\log x}{x^{2}}=0 \\ &\Rightarrow 1-\log x=0, x \neq 0 \\ &\Rightarrow \log x=1 \\ &x=e^{1}=e \end{aligned}$
Differentiating $\frac{d y}{d x}$ with respect to $x$
$\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}} &=\frac{\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(1-\log \mathrm{x})-(1-\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{2}\right)}{\mathrm{x}^{4}} . \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{\mathrm{x}^{2}\left(-\frac{1}{\mathrm{x}}\right)-2 \mathrm{x}(1-\log \mathrm{x})}{\mathrm{x}^{4}} \\ &=\frac{-\mathrm{x}-2 \mathrm{x}+2 \mathrm{xlog} \mathrm{x}}{\mathrm{x}^{4}} \\ &=\frac{-3 \mathrm{x}-2 \mathrm{x} \log \mathrm{x}}{\mathrm{x}^{4}} \end{aligned}$
At x=e,
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-3 e+2 e \log e}{e^{4}} \\ &=\frac{-3 e+2 e(1)}{e^{4}} \ldots \log e=1 \\ &=-\frac{e}{e^{4}}=-\frac{1}{e^{3}}<0 \end{aligned}$
So x = e is a point of local maxima

Maxima and Minima exercise 17.3 question 5

Answer:
x = 0 is local minima and its minimum value is 2 and x =1 is local maxima and its maximum value is -6
Given:
$f(x)=\frac{4}{x+2}+x$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(4(\mathrm{x}+2)^{-1}+\mathrm{x}\right) \\ &=-4(\mathrm{x}+2)^{-2}+1 \\ &=-\frac{4}{(x+2)^{2}}+1 \\ &=\frac{-4+\mathrm{x}^{2}+4 \mathrm{x}+4}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}(\mathrm{x}+4)}{(\mathrm{x}+2)^{2}} \end{aligned}$
$\begin{aligned} &\text { Put } f(x)=0 \\ &\frac{x(x+4)}{(x+2)^{2}}=0 \\ &x(x+4)=0 \\ &\text { i.e } x=0, \text { or } \\ &x=-4 \end{aligned}$
Thus x = 0 and x = -4 are the points of local maxima and minima
Differentiating f’ with respect to x
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right) \\ &=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{(\mathrm{x}+2)^{2}(2 \mathrm{x}+4)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \cdot 2(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left[(\mathrm{x}+2)^{2}-\left(\mathrm{x}^{2}+4 \mathrm{x}\right)\right]}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left(\mathrm{x}^{2}+4 \mathrm{x}+4-\mathrm{x}^{2}-4 \mathrm{x}\right)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8}{(\mathrm{x}+2)^{3}} \end{aligned}$
Put x = 0 in f ‘’ (x)
$\begin{aligned} \mathrm{f}^{\prime \prime}(0) &=\frac{8}{(0+2)^{3}} \\ &=\frac{8}{8} \\ &=1>0 \end{aligned}$
S,x=0is local minima. Hence,its minimum value will be f(0)=2
And x= -4 in f ‘’(x)
$\begin{aligned} &f^{\prime \prime}(-4)=\frac{8}{(-4+2)^{3}} \\ &=\frac{8}{-8} \\ &=-1<0 \end{aligned}$
So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6

Maxima and Minima exercise 17.3 question 6

Answer:
$x=\frac{\pi}{4}$ is a point of local minima and its local minimum value will be $f\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
$x=\frac{3\pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3\pi}{4}\right)=-1-\frac{3\pi}{2}$
Hint:
$\text { Put } f^{\prime} x=0$
Given:
$f(x)=\tan x-2 x$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} &f(x)=\sec ^{2} x-2\\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0\\ &\sec ^{2} x-2=0\\ &\Rightarrow \sec ^{2} \mathrm{x}=2\\ &\Rightarrow \sec \mathrm{x}=\pm \sqrt{2}\\ &=\sec x=\sqrt{2}\\ &\Rightarrow \mathrm{x}=\frac{\pi}{4}\\ &\text { or, } \sec x=-\sqrt{2}\\ &\text { or, } \sec \mathrm{x}=\pi-\frac{\pi}{4}\\ &=\frac{3 \pi}{4} \end{aligned}$
Thus $x=\frac{\pi}{4}$ or $x=\frac{3\pi}{4}$ is possible points of local maxima and minima
Differentiating f’(x) with respect to x
$\begin{aligned} &\mathrm{f}^{\mathrm{u}}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec ^{2} \mathrm{x}-2\right] \\ &=2 \sec \mathrm{x}(\sec \mathrm{xtan} \mathrm{x}) \\ &=2 \sec ^{2} \mathrm{x} \tan \mathrm{x} \\ &\text { Put } \mathrm{x}=\frac{\pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x}) \\ &\begin{aligned} \mathrm{f}=\left(\frac{\pi}{4}\right) &=2 \sec ^{2}\left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{4}\right) \\ &=2(2)(1) \\ &=4>0 \end{aligned} \end{aligned}$
So ,
$\mathrm{x}=\frac{\pi}{4}$ is a point of Local Minima and its local maximum value will be $\mathrm{f}\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
$\begin{aligned} &\text { And Put, }\\ &\mathrm{x}=\frac{3 \pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})\\ &\mathrm{f}^{\prime \prime}\left(\frac{3 \pi}{4}\right)=2 \sec ^{2}\left(\frac{3 \pi}{4}\right) \tan \left(\frac{3 \pi}{4}\right)\\ &=2(-\sqrt{2})^{2} \quad(-1)\\ &=-4<0 \end{aligned}$
So $x=\frac{3 \pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3 \pi}{4}\right)=-1-\frac{3 \pi}{2}$

Maxima and Minima exercise 17.3 question 7
Answer:
$\begin{aligned} &a=3 \\ &b=-9 \& c \in R \end{aligned}$
Hint:
Put,
$\begin{aligned} &\mathrm{f}^{\prime}(-1)=0 \\ &\mathrm{f}^{\prime}(3)=0 \end{aligned}$
Given:
$f(x)=x^{3}+a x^{2}+b x+C$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} f^{\prime}(x) &=3 x^{2}+2 a x+b+0 \\ &=3 x^{2}+2 a x+b \end{aligned}$
$\begin{aligned} &\text { Calculate } \mathrm{f}^{\prime}(\mathrm{x}) \text { at } \mathrm{x}=-1 \text { and } \mathrm{x}=3\\ &\begin{aligned} \mathrm{f}^{\prime}(-1) &=3(-1)^{2}+2 \mathrm{a}(-1)=\mathrm{b} \\ &=3-2 \mathrm{a}+\mathrm{b} \\ \mathrm{f}^{\prime}(3) &=3(3)^{2}+2 \mathrm{a}(3)+\mathrm{b} \\ &=27+6 \mathrm{a}+\mathrm{b} \end{aligned} \end{aligned}$
Put and $f^{\prime}(-1) =0$ and $f^{\prime}(3)=0$ as they are the extremum values hence $f^{\prime}(x)=0$
$\begin{aligned} &\therefore 3-2 \mathrm{a}+\mathrm{b}=0 \\ &7-2 \mathrm{a}+\mathrm{b}=-3-(1) \\ &27-6 \mathrm{a}+\mathrm{b}=20 \mid \\ &6 \mathrm{a}+\mathrm{b}=-27-(2) \\ &(1)-(2), \mathrm{wc} \text { get } \\ &-2 \mathrm{a}-6 \mathrm{a}=-3+27 \\ &-8 \mathrm{a}=2 \mathrm{y} \\ &\mathrm{a}=3 \\ &\text { Put } \mathrm{a}=3 \text { in }(1) \\ &-2(3)^{2}-\mathrm{b}=-3 \\ &\mathrm{~b}=-9 \end{aligned}$

Maxima and Minima exercise 17.3 question 8

Answer:
Hence proved.
Hint: $\mathrm{f}^{\prime}(\mathrm{x})=0$
Put,
Given:
$\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\sqrt{3} \cos \mathrm{x}$
Explanation:
Differentiating fx with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\cos x-\sqrt{3} \sin \mathrm{x} \\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0 \\ &\cos \mathrm{x}-\sqrt{3} \sin \mathrm{x}=0 \\ &\Rightarrow \sqrt{3} \sin \mathrm{x}=\cos \mathrm{x} \\ &\Rightarrow \tan \mathrm{x}=\frac{1}{\sqrt{3}} \end{aligned}$
$\Rightarrow x=\frac{\pi}{6}$
Differentiating f’(x) with respect to x ,
$\begin{aligned} &f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x \\ &\text { Putt } x=\frac{\pi}{6} \text { at } f^{\prime \prime}(x) \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6} \\ &=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ &=-\frac{1}{2}-\frac{3}{2} \\ &=-\frac{4}{2} \\ &=-2<0 \end{aligned}$
So, $x = \frac{\pi}{6}$ is a point local maxima.

Rd Sharma Class 12th exercise 17.3 is an essential part of the RD Sharma solutions for maths. The Class 12 RD Sharma chapter 17 exercise 17.3 solution will contain an important chapter of the NCERT maths book that needs to be studied well. RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 is going to be based on the chapter Maxima and minima.
Rd Sharma class 12-chapter 17 exercises 17.3 answers that are basic and simple. It will help students learn all formulae well and become skilled at the subject in general. RD Sharma class 12 solutions ex 17.3 has around 21 inquiries.

Solutions identified with the Greatest and least upsides of a function in its space
Definition and which means of greatest
Definition and significance of least
Nearby maxima
Questions identified with Nearby minima
Discover Solutions of First subordinate test for neighborhood maxima and minima
Higher-request subordinate test

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If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi.

3 Jobs Available

Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

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Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

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Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

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Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

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QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

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Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available

Safety Manager

2 Jobs Available

Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

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Career as IT Manager requires managing the various aspects of an organization's information technology systems. He or she is responsible for increasing productivity and solving problems related to software and hardware. While this role is typically one of the lower-level positions within an organisation, it comes with responsibilities related to people and ownership of systems.

2 Jobs Available